What is the integral from 0 to 2pi of 7x^2sin(1/4 x) ?

The integral from 0 to 2pi of 7x^2sin(1/4 x) is 448(pi-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2pi of 7x^2sin(1/4 x)

\int_{0}^{2 π} 7 x^{2} sin (\frac{1}{4} x) d x

448 (π - 2)

Decimal Notation

511.43350 \dots

Solution steps

\int_{0}^{2 π} 7 x^{2} sin (\frac{1}{4} x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot \int_{0}^{2 π} x^{2} sin (\frac{1}{4} x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} 64 u^{2} sin (u) d u

= 7 \cdot \int_{0}^{\frac{π}{2}} 64 u^{2} sin (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 7 \cdot 64 \cdot \int_{0}^{\frac{π}{2}} u^{2} sin (u) d u

Apply Integration By Parts

: [- u^{2} cos (u) - \int - 2 u cos (u) d u]_{0}^{\frac{π}{2}}

= 7 \cdot 64 [- u^{2} cos (u) - \int - 2 u cos (u) d u]_{0}^{\frac{π}{2}}

\int - 2 u cos (u) d u = - 2 (u sin (u) + cos (u))

= 7 \cdot 64 [- u^{2} cos (u) - (- 2 (u sin (u) + cos (u)))]_{0}^{\frac{π}{2}}

Simplify

= 448 [- u^{2} cos (u) + 2 (u sin (u) + cos (u))]_{0}^{\frac{π}{2}}

Compute the boundaries:

π - 2

= 448 (π - 2)

\frac{d y}{d x} = \frac{( y - x )}{x}

n = 2 \sum \infty - \frac{1}{4 ^{n}}

x \to 0 lim (\frac{9 ^{x} - 5 ^{x}}{x})

\frac{\partial}{\partial y} (\frac{y ln ( y )}{x - 1})

\int \frac{2}{x 9 + x ^{2}} d x

What is the integral from 0 to 2pi of 7x^2sin(1/4 x) ?
The integral from 0 to 2pi of 7x^2sin(1/4 x) is 448(pi-2)